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freemind
Riemann Hypothesis
Riemann Hypothesis


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Joined: 14 Jul 2004
Posts: 332
Location: MIT or Moldova
Moldova, Republic ofUnited States

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#1
Nice: find number of even permutations with no fixed points
Moldova 2008 IMO-BMO Second TST Problem 4

Find the number of even permutations of \{1,2,\ldots,n\} with no fixed points.
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The fate of equilibrium is to end the eternity...

PostPosted: Sat Mar 29, 2008 4:49 pm
pohoatza
Navier-Stokes Equations
Navier-Stokes Equations

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Joined: 27 Apr 2006
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Location: Bucharest
Romania

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#2
It is known that an even permutation of this kind is called an even derrangement. In the following, for sake of clarity, denote by O_{n}, E_{n}, and D_{n} the numbers of odd, even, and all derangements of the n numbers. Now, I shall figure out a connection between O_{n} and E_{n}, the conclusion following then from the obvious fact that O_{n} + E_{n} = D_{n}.

First, it is clear that E_{1} = O_{1} = 0, E_{2} = 0, O_{2} = 1. From the initial ordering of the n numbers, interchange the first and the j-th numbers. There will be E_{n - 2} derangements of the remaining n - 2 letters which lead to odd derangements of the entire set. Since j may be any one of the integers 2, 3, \ldots , n, there are (n - 1)E_{n - 2} odd derangements of this type. Again, starting with the original arrangement of the n integers, consider the first one followed by any one of the E_{n - 1} even derangements of the last n - 1 numbers. Now interchange the two numbers from the first and j-th positions, and we obtain an odd derangement of all n numbers. There are (n - 1)E_{n - 1} odd derangements of this type. Thus, O_{n} = (n - 1)(E_{n - 1} + E_{n - 1}), and likewise, E_{n} = (n - 1)(O_{n - 1} + O_{n - 2}).
In conclusion, by a small induction, one can deduce that O_{n} - E_{n} = ( - 1)^{n}(n - 1), and since O_{n} + E_{n} = D_{n}, we have that E_{n} = \frac {1}{2}(D_{n} + ( - 1)^{n-1}(n - 1)), where D_{n} = n!\left(1 - \frac {1}{1!} + \ldots + \frac {( - 1)^{n}}{n!}\right), as a consequence, for example, of the Inclusion-Exclusion Principle.

I suspect that my approach is far from beeing new, but this is what I've found. I am also aware of a solution using some linear algebra, from "Recounting the Odds of an Even Derangement", Mathematics Magazine.
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Cosmin Pohoata, Bucharest, Romania

PostPosted: Sat Mar 29, 2008 8:42 pm
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